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((cos^3(a)))/((2cos^2(a)-1))=cos(a)

Solution

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

Solution

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = π + 2 πn, a = 2 πn

Degrees

a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

Solve by substitution

\frac{cos ^{3} ( a )}{2 cos ^{2} ( a ) - 1} = cos (a)

Let:

cos (a) = u

\frac{u ^{3}}{2 u ^{2} - 1} = u

\frac{u ^{3}}{2 u ^{2} - 1} = u : u = 0, u = - 1, u = 1

\frac{u ^{3}}{2 u ^{2} - 1} = u

Multiply both sides by

2 u^{2} - 1

\frac{u ^{3}}{2 u ^{2} - 1} = u

Multiply both sides by

2 u^{2} - 1

\frac{u ^{3}}{2 u ^{2} - 1} (2 u^{2} - 1) = u (2 u^{2} - 1)

Simplify

u^{3} = u (2 u^{2} - 1)

u^{3} = u (2 u^{2} - 1)

Solve

u^{3} = u (2 u^{2} - 1) : u = 0, u = - 1, u = 1

u^{3} = u (2 u^{2} - 1)

Expand

u (2 u^{2} - 1) : 2 u^{3} - u

u (2 u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = u, b = 2 u^{2}, c = 1

= u \cdot 2 u^{2} - u \cdot 1

= 2 u^{2} u - 1 \cdot u

Simplify

2 u^{2} u - 1 \cdot u : 2 u^{3} - u

2 u^{2} u - 1 \cdot u

2 u^{2} u = 2 u^{3}

2 u^{2} u

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= 2 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 2 u^{3}

1 \cdot u = u

1 \cdot u

Multiply:

1 \cdot u = u

= u

= 2 u^{3} - u

= 2 u^{3} - u

u^{3} = 2 u^{3} - u

Switch sides

2 u^{3} - u = u^{3}

Move

u^{3}

to the left side

2 u^{3} - u = u^{3}

Subtract

u^{3}

from both sides

2 u^{3} - u - u^{3} = u^{3} - u^{3}

Simplify

u^{3} - u = 0

u^{3} - u = 0

Factor

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

Factor out common term

u : u (u^{2} - 1)

u^{3} - u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{3} = u^{2} u

= u^{2} u - u

Factor out common term

u

= u (u^{2} - 1)

= u (u^{2} - 1)

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

u (u + 1) (u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

The solutions are

u = 0, u = - 1, u = 1

u = 0, u = - 1, u = 1

Verify Solutions

Find undefined (singularity) points:

u = \frac{1}{2}, u = - \frac{1}{2}

Take the denominator(s) of

\frac{u ^{3}}{2 u ^{2} - 1}

and compare to zero

Solve

2 u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{2} - 1 = 0

Move

1

to the right side

2 u^{2} - 1 = 0

Add

1

to both sides

2 u^{2} - 1 + 1 = 0 + 1

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

Simplify

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

\frac{1}{2} = \frac{1}{2}

\frac{1}{2}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{2}

- \frac{1}{2} = - \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

The following points are undefined

u = \frac{1}{2}, u = - \frac{1}{2}

Combine undefined points with solutions:

u = 0, u = - 1, u = 1

Substitute back

u = cos (a)

cos (a) = 0, cos (a) = - 1, cos (a) = 1

cos (a) = 0, cos (a) = - 1, cos (a) = 1

cos (a) = 0 : a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

cos (a) = 0

General solutions for

cos (a) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

General solutions for

cos (a) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = π + 2 πn

a = π + 2 πn

cos (a) = 1 : a = 2 πn

cos (a) = 1

General solutions for

cos (a) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 0 + 2 πn

a = 0 + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = π + 2 πn, a = 2 πn

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